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Section 5.1 The Adjoint Representation

Any Lie algebra \(\gg\) acts on itself via commutators. This action is linear, so we can represent it using matrices.

Consider first \(\su(2)\text{.}\) We know that

\begin{equation} \left[\frac12 s_p, \frac12 s_q\right] = \epsilon_{pqr} \left(\frac12 s_r\right) ,\tag{5.1.1} \end{equation}

where \(\epsilon_{pqr}\) is completely antisymmetric in its indices and satisfies \(\epsilon_{123}=1\text{.}\) Taking our basis to be \(\{\frac12 s_p\}\) for \(p=1,2,3\text{,}\) we can work out the matrix representation of \(s_p\text{.}\) For instance, since \(\frac12 s_1\) takes \(\frac12 s_2\) to \(\frac12 s_3\) and \(\frac12 s_3\) to \(-\frac12 s_2\) (and of course commutes with itself), the matrix corresponding to \(\frac12 s_1\) is

\begin{equation} \frac12 S_1 = \begin{pmatrix} 0 \amp 0 \amp 0 \\ 0 \amp 0 \amp -1 \\ 0 \amp 1 \amp 0 \\ \end{pmatrix}\tag{5.1.2} \end{equation}

in this basis. The matrices for \(\frac12 s_2\) and \(\frac12 s_3\) can be obtained similarly, or by cyclic permutations.

It is straightforward to verify that the adjoint representation is indeed a representation, that is, that the map

\begin{equation} \textrm{ad}: \frac12 s_p \longmapsto \frac12 S_p\tag{5.1.3} \end{equation}

a Lie algebra homomorphism. Yes, we could have dispensed with the factors of \(\frac12\) throughout, but you should recognize the matrices \(\frac12 S_m\) as being the basis \(r_m\) that we introduced for \(\so(3)\text{.}\) Thus,

\begin{equation} \textrm{ad}(\su(2)) \equiv \so(3)\tag{5.1.4} \end{equation}

where we have written \(\equiv\) to emphasize that this relationship is an equality of Lie algebras, rather than an isomorphism.

For a less obvious example, consider \(\sl(2,\RR)\text{,}\) with its standard basis \(\{\sigma_0,\sigma_\pm\}\text{.}\) We know that

\begin{equation} [\sigma_0,\sigma_\pm] = \pm \sigma_\pm\tag{5.1.5} \end{equation}

and of course \(\sigma_0\) commutes with itself, so the matrix of \(\sigma_0\text{,}\) acting via commutators, is in this basis

\begin{equation} \Sigma_0 = \begin{pmatrix} 0 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \\ 0 \amp 0 \amp -1 \\ \end{pmatrix}\tag{5.1.6} \end{equation}

which, not surprisingly, is diagonal. Similarly, the matrices of \(\sigma_\pm\) are

\begin{equation} \Sigma_+ = \begin{pmatrix} 0 \amp 0 \amp 2 \\ -1 \amp 0 \amp 0 \\ 0 \amp 0 \amp 0 \\ \end{pmatrix}, \qquad \Sigma_- = \begin{pmatrix} 0 \amp -2 \amp 0 \\ 0 \amp 0 \amp 0 \\ 1 \amp 0 \amp 0 \\ \end{pmatrix} ,\tag{5.1.7} \end{equation}

since \([\sigma_+,\sigma_-]=2\sigma_0\text{.}\) Again, it is easily checked that the commutators are preserved, that is, that

\begin{equation} [\Sigma_0,\Sigma_\pm] = \pm \Sigma_\pm , \qquad [\Sigma_+,\Sigma_-]=2\Sigma_0 .\tag{5.1.8} \end{equation}

Matrix Lie algebras such as \(\su(n)\) are normally defined in terms of the properties of certain matrices. Since these matrices have—by definition—the appropriate commutators, they form a representation of the Lie algebra known as the defining representation. Another commonly used representation is the minimal representation, which consists of the smallest matrices that reproduce the given commutators.

For example, the defining representation of \(\su(2)\) acts on \(\CC^2\text{,}\) whereas the defining representation of \(\so(3)\) acts on \(\RR^3\text{.}\) Even though \(\su(2)\cong\so(3)\text{,}\) both of these representations are referred to as minimal, although correct usage would distinguish whether the representation is real or complex. The adjoint representation in both cases is, of course, 3-dimensional. A more interesting example is \(\su(3)\text{,}\) where the defining representation is again minimal and is 3-dimensional, and the adjoint representation is 8-dimensional. In this context, “dimension” refers to the size of the matrices, or equivalently to the dimension of the vector space being acted on.

Strictly speaking, the Killing form is always defined in the adjoint representation, that is,

\begin{equation} B(X,Y) := \tr\bigl( \textrm{ad}(X)\,\textrm{ad}(Y) \bigr) ,\tag{5.1.9} \end{equation}

although in matrix Lie algebras computing the trace directly with \(X\) and \(Y\) leads to an inner product that differs at most by an overall scale.